Converging or diverging 1/ln(n)

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In summary, the conversation discusses the attempt to determine if the series n=2 to inf. of 1/ln(n) converges or diverges using various tests such as the limit test, integral test, and ratio test. The individual is struggling to find a method that works and considers using the direct comparison test with the harmonic series. After realizing a mistake in their logic, they conclude that the series must diverge.
  • #1
BoldKnight399
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Determine if the series n=2 to inf. of 1/ln(n) converges or diverges


Ok so first I tried the limit test (the simple one) and found that it was 0 which was not helpful at all. Then I tried the integral test. It came out to be (integral)1/ln(n)=n/ln(n) + n/(ln(n))^2 + 2(integral from 2 to infin.) 1/(ln(n))^3. I was thinking of possibly doing a direct comparison test, but I have no clue what to compare it to. So then I tried the ratio test. That also failed, because the limit of the absolute value of the ratio was equal to 1 thus inconclusive and leaving me back where I started.

I have no idea how else to approach this problem. I am hoping that I maybe just messed up my integration. If anyone has a clue how to approach this that would be great.
 
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  • #2
Try comparing it with the harmonic series
 
  • #3
ahhhh. Just to check. When I do a direct comparison to the harmonic series (which diverges) 1/ln(n) is larger so it must also converge. Is this right or have I been staring at this problem long enough that my logic is swiss cheese?
 
  • #4
Check the criteria for the comparison test again, you've made one mistake.
 
  • #5
The only criteria I have for the Direct Comparison test is that "if a series is less than or equal to a converging series then it also converges." and "if it is greater than a diverging series it also diverges." Am I missing something?
 
  • #6
o wait...duh since harmonic diverges and the 1/lnn is larger it must diverge. that was stupid on my part
 

FAQ: Converging or diverging 1/ln(n)

What is the significance of "Converging or diverging 1/ln(n)" in mathematics?

"Converging or diverging 1/ln(n)" is a mathematical expression used to determine the behavior of a series, which is a sum of infinitely many terms. It helps to determine if the series will approach a finite value (converge) or go to infinity (diverge) as the number of terms increases.

How do you calculate the convergence or divergence of 1/ln(n)?

To calculate the convergence or divergence of 1/ln(n), you can use the Integral Test, which compares the series to an integral expression. If the integral converges, then the series converges and if the integral diverges, then the series diverges. Alternatively, you can also use the Comparison Test or the Limit Comparison Test to determine the behavior of the series.

Can the series 1/ln(n) ever converge?

Yes, the series 1/ln(n) can converge under certain conditions. For example, if the series is modified to become 1/ln(n+1), then it will converge. Additionally, if the series is multiplied by a constant, such as 2/ln(n), then it will also converge. However, in its original form, the series diverges.

How is the behavior of 1/ln(n) related to the natural logarithm function?

The behavior of 1/ln(n) is closely related to the natural logarithm function because the denominator, ln(n), represents the rate of growth of the series. As n increases, ln(n) also increases, which causes the series to approach infinity (diverge). This relationship is important in determining the convergence or divergence of the series.

Are there any real-world applications of "Converging or diverging 1/ln(n)"?

Yes, there are several real-world applications of "Converging or diverging 1/ln(n)". One example is in finance, where it is used to determine the rate of return on investments over time. It is also used in physics and engineering to analyze the behavior of systems with infinitely many components. Additionally, it is used in computer science to analyze the efficiency of algorithms and data structures.

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